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Find values of constants a and b such that the given improper integral converges

by RoRi

Find values for real numbers a and b so that the following improper integral equation holds:

    \[ \int_1^{\infty} \left( \frac{2x^2 + bx + a}{x(2x+a)} - 1 \right) \, dx = 1. \]

First, we have

    \begin{align*} \int_1^{\infty} \left( \frac{2x^2 + bx + a}{x(2x+a)} - 1 \right) \, dx &= \int_1^{\infty} \frac{2x^2 + bx + a - 2x^2 - ax}{2x^2+ax} \, dx \\[9pt] &= \int_1^{\infty} \frac{(b-a)x + a}{2x^2 + ax} \, dx. \end{align*}

In order for this to converge we must have the coefficient of x in the numerator equal to 0; hence, we must have a = b. Now, to solve the integral equation we have

    \begin{align*} \int_1^{\infty} \frac{a}{2x^2 + ax} \,dx &= \int_1^{\infty} \frac{a}{x(2x+a)} \, dx \\[9pt] &= \int_1^{\infty} \left( \frac{1}{x} - \frac{2}{2x+a} \right) \, dx &(\text{partial fractions})\\[9pt] &= \lim_{c \to \infty} \int_1^c \left( \frac{1}{x} - \frac{2}{2x+a} \right) \, dx \\[9pt] &= \lim_{c \to \infty} \left( \log |x| - \log |2x+a| \right)\Bigr \rvert_1^c \\[9pt] &= \lim_{c \to \infty} \left( \log \left| \frac{x}{2x+a} \right| \right) \Bigr \rvert_1^c \\[9pt] &= \lim_{c \to \infty} \left( \log \left( \frac{c}{2c+a} \right) - \log \left( \frac{1}{2+a} \right) \right) \\[9pt] &= \log \frac{1}{2} - \log \frac{1}{2+a} \\[9pt] &= \log \frac{2+a}{2}. \end{align*}

But, by assumption this integral equals 1, so we have

    \begin{align*} \log \frac{2+a}{2} = 1 && \implies && \frac{2+a}{2} &= e \\[9pt] && \implies && 2+a &= 2e \\[9pt] && \implies && a & =2e - 2. \end{align*}

Since b = a we then have

    \[ a = b = 2e - 2. \]

Find a value of C so that the given improper integral converges

by RoRi

Find a value for the constant C \in \mathbb{R} so that the improper integral

    \[ \int_0^{\infty} \left( \frac{1}{\sqrt{1+x^2}} - \frac{C}{x+1} \right) \, dx \]

converges and find the value of the integral in this case.

First, we have

    \begin{align*} \int_0^{\infty} \left( \frac{1}{\sqrt{1+2x^2}} - \frac{C}{x+1} \right) \, dx &= \int_0^{\infty} \frac{x+1-C\sqrt{1+2x^2}}{(x+1)\sqrt{1+2x^2}} \, dx \\[9pt] &= \int_0^{\infty} \frac{x+1-xC\sqrt{2 + \frac{1}{x^2}}}{(x^2+x)\sqrt{2 + \frac{1}{x^2}}} \, dx \\[9pt] &= \int_0^{\infty} \frac{x \left(1 - C \sqrt{2 + \frac{1}{x^2}} \right) + 1}{x^2 \sqrt{2 + \frac{1}{x^2}} + x \sqrt{2 + \frac{1}{x^2}}} \, dx. \end{align*}

For this to converge we must have the coefficient of the x in the numerator going to 0 as x \to \infty (otherwise the integral will converge by limit comparison with \int \frac{1}{x}). Hence,

    \[ \lim_{x \to \infty} \left(1 - C \sqrt{2 + \frac{1}{x^2}} \right) = 1 - \sqrt{2} C \quad \implies \quad C = \frac{\sqrt{2}}{2}. \]

Now, we need to evaluate the integral. Incomplete.

Find a value of the constant C so that the given improper integral converges

by RoRi

Find the value of C \in \mathbb{R} so that the improper integral

    \[ \int_1^{\infty} \left( \frac{x}{2x^2+2C} - \frac{C}{x+1} \right) \, dx \]

converges, and compute the value of the integral in this case.

First, we compute the value of the constant C,

    \begin{align*} \int_1^{\infty} \left( \frac{x}{2x^2+2C} - \frac{C}{x+1} \right) \, dx &= \int_1^{\infty} \frac{x^2+x-2C(x^2+C)}{2(x^2+C)(x+1)} \, dx \\[9pt] &= \int_1^{\infty} \frac{(1-2C)x^2 + x - 2C^2}{2(x^2+C)(x+1)} \, dx. \end{align*}

This integral will diverge by comparison with \int_1^{\infty} \frac{1}{x} \, dx if the coefficient of the x^2 term in the numerator is not equal to zero. Hence, we must have

    \[ 1-2C = 0 \quad \implies \quad C = \frac{1}{2}. \]

Next, we evaluate the integral with C = \frac{1}{2},

    \begin{align*} \int_1^{\infty} \left( \frac{x}{2x^2+2C} - \frac{C}{x+1} \right) \, dx &= \int_1^{\infty} \left( \frac{x}{2x^2+1} - \frac{1}{2(x+1)} \right) \, dx \\[9pt] &= \lim_{a \to +\infty} \int_1^a \left( \frac{x}{2x^2+1} - \frac{1}{2(x+1)} \right) \, dx \\[9pt] &= \lim_{a \to +\infty} \left( \frac{1}{4} \int_1^a \frac{4x}{2x^2+1} \, dx - \frac{1}{2} \int_1^a \frac{1}{x+1} \, dx \right) \\[9pt] &= \lim_{a \to +\infty} \left( \frac{1}{4} \log(2x^2+1) - \frac{1}{2} \log |x+1| \right) \Bigr \rvert_1^a \\[9pt] &= \frac{1}{4} \cdot \lim_{a \to +\infty} \left( \log(2a^2+1) - \log 3 - 2 \log (a+1) + 2 \log 2 \right) \\[9pt] &= \frac{1}{4} \cdot \lim_{a\to +\infty} \left( \log \frac{2a^2+1}{(a+1)^2} + \log \frac{4}{3} \right) \\[9pt] &= \frac{1}{4} \cdot \log \frac{8}{3}. \end{align*}

Test the improper integral ∫ 1 / x (log x)s for convergence

by RoRi

Test the following improper integral for convergence:

    \[ \int_2^{\infty} \frac{dx}{x (\log x)^s}. \]

The integral converges if s < 1 and diverges for s \geq 1.

Proof. We can compute this directly. We evaluate the indefinite integral (using the substitution u = \log x) for s \neq 1:

    \begin{align*} \int \frac{1}{x (\log x)^s} \, dx &= \int \frac{1}{u^s} \, du \\ &= \frac{u^{1-s}}{1-s} \\ &= \frac{(\log x)^{1-s}}{1-s}. \end{align*}

But then the limit

    \[ \lim_{x \to +\infty} \frac{(\log x)^{1-s}}{1-s} \]

is finite for s > 1 and diverges for s \leq 1 (since \log x \to \infty as x \to \infty and so the limit diverges when 1-s > 0 and converges for 1-s < 0). Hence, the integral

    \[ \int_2^{\infty} \frac{dx}{x (\log x)^s} \]

converges for s > 1 and diverges for s < 1. In the case that s = 1 the indefinite integral is

    \[ \int \frac{1}{x \log x} \, dx = \log (\log x) \]

and \log (\log x) \to +\infty as x \to +\infty, so the integral divers for s = 1 as well. Therefore, we have the integral converges for s > 1 and diverges for s \leq 1. \qquad \blacksquare

Test the improper integral ∫ 1 / (x1/2 log x) for convergence

by RoRi

Test the following improper integral for convergence:

    \[ \int_{0^+}^{1^-} \frac{dx}{\sqrt{x} \log x}. \]

The integral diverges.

Proof. First, we show that \log x < \sqrt{x} for all x> 0. To do this let

    \[ f(x) = \sqrt{x} - \log x \quad \implies \quad f'(x) = \frac{1}{2 \sqrt{x}} - \frac{1}{x} = \frac{\sqrt{x} - 2}{2x}. \]

This derivative is 0 at x = 4 and is less than 0 for x < 4 and greater than 0 for x > 4. Hence, f(x) has a minimum at x = 4. But, f(4) = \sqrt{4} - \log 4 = 2 - 2 \log 2 > 0 (since 2 < e implies \log 2 < 1). So, f(x) has a minimum at x = 4 and is positive there; thus, it is positive for all x > 0, or

    \[ f(x) = \sqrt{x} - \log x > 0 \quad \implies \quad \sqrt{x} > \log x \qquad \text{for all } x > 0. \]

So, since \log x < \sqrt{x} we know

    \[ \frac{1}{\sqrt{x} \log x} > \frac{1}{\sqrt{x} \cdot \sqrt{x}} = \frac{1}{x}. \]

But then, consider the limit

    \begin{align*} \lim_{x \to 0^+} \frac{ \frac{1}{x} }{ \frac{1}{\sqrt{x} \log x}} &= \lim_{x \to 0^+} \frac{\log x}{\sqrt{x}} \\ &= \lim_{x \to 0^+} \frac{\log x}{x^{\frac{1}{2}}} \\[9pt] &= 0 &(\text{by Theorem 7.11}). \end{align*}

Therefore, by the limit comparison test (Theorem 10.25), the convergence of \frac{1}{\sqrt{x} \log x} would imply the convergence of \int_{0^+}^a \frac{1}{x} (for any 0 < a < 1), but we know by Example 5 that this integral diverges. Hence, we must also have the divergence of

    \[ \int_{0^-}^{1^+} \frac{1}{\sqrt{x} \log x} \, dx. \qquad \blacksquare \]

Test the improper integral ∫ log x / (1-x) for convergence

by RoRi

Test the following improper integral for convergence:

    \[ \int_{0^+}^{1^-} \frac{\log x}{1-x} \, dx. \]

The integral converges.

Proof. First, we write

    \[ \int_{0^+}^{1^-1} \frac{\log x}{1-x} \,dx = \int_{0^+}^{\frac{1}{2}} \frac{\log x}{1-x} \, dx + \int_{\frac{1}{2}}^{1^-} \frac{\log x}{1-x} \, dx. \]

For the first integral we know for all x \in \left( 0 , \frac{1}{2} \right) we have \frac{1}{2} \leq (1-x) < 1; hence,

    \[ \frac{\log x}{1-x} < 2 \log x \qquad \text{for all } x \in \left( 0, \frac{1}{2} \right). \]

Then, the integral

    \begin{align*} 2 \int_{0^+}^{\frac{1}{2}} \log x \, dx &= 2 \cdot \lim_{a \to 0^+} \int_a^{\frac{1}{2}} \log x \, dx \\[9pt] &= 2 \cdot \lim_{a \to 0^+} \left( x \log x - x \right)\Bigr \rvert_a^{\frac{1}{2}} \\[9pt] &= 2 \cdot \lim_{a \to 0^+} \left( \frac{1}{2} \log \frac{1}{2} - \frac{1}{2} - a \log a + a \right) \\[9pt] &= \log \frac{1}{2} - 1 \\[9pt] &= - \log 2 - 1. \end{align*}

(We know \lim_{a \to 0} x \log x = 0 by L’Hopital’s, writing x \log x = \frac{\log x}{1/x} or by Example 2 on page 302 of Apostol.) Hence,

    \[ \int_{0^+}^{\frac{1}{2}} \frac{\log x}{1-x} \, dx \]

converges by the comparison theorem (Theorem 10.24 on page 418 of Apostol).

For the second integral, we use the expansion of \log x about x = 1,

    \[ \log x = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \cdots. \]

Then we have

    \begin{align*} \int_{\frac{1}{2}}^{1^-} \frac{\log x}{1-x} \, dx &= \int_{\frac{1}{2}}^{1^-} \frac{ (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \cdots }{1-x} \, dx \\[9pt] &= \int_{\frac{1}{2}}^{1^-} \left( -1 + \frac{x-1}{2} - \frac{(x-1)^2}{3} + \cdots \right) \, dx. \end{align*}

But this integral converges since it has no singularities.

Thus, we have established the convergence of

    \[ \int_{0^+}^{1^-} \frac{\log x}{1-x} \, dx. \qquad \blacksquare \]